texstuff

Author: cedrict

basis_P1_2D.tex

@@ -9,13 +9,21 @@
 Velocities (or displacements) $(u^h,v^h)$ in the element are interpolated from nodal velocities
 $(u_i,v_i)$ using basis functions $\bN_i$ as follows,
 \[
+\vec\upnu^h=
 \left(
 \begin{array}{c}
 u^h(x,y) \\v^h(x,y)
 \end{array}
 \right)
 =
 \left(
+\begin{array}{c}
+\sum_{i=1}^3 \bN_i(x,y) u_i \\
+\sum_{i=1}^3 \bN_i(x,y) v_i
+\end{array}
+\right)
+=
+\left(
 \begin{array}{cccccc}
 \bN_1(x,y) & 0 & \bN_2(x,y) & 0 & \bN_3(x,y) & 0\\
 0 & \bN_1(x,y) & 0 & \bN_2(x,y) & 0 & \bN_3(x,y)\\
@@ -29,16 +37,10 @@
 \right)
 \]
 or simply 
-\[
-u^h(x,y)=\sum_{i=1}^3 \bN_i(x,y) u_i
-\qquad
-\qquad
-v^h(x,y)=\sum_{i=1}^3 \bN_i(x,y) v_i
-\]
 
 For this element, we have three nodes at the vertices of the triangle, which are 
 numbered around the element in the counterclockwise direction. 
-Each node has two degrees of freedom (can move in the $x$ and $y$ directions). 
+Each node has two degrees of freedom (i.e. it can move in the $x$ and $y$ directions). 
 The velocities $u^h$ and $v^h$ are assumed to be linear functions within the element, that is, 
 \begin{eqnarray}
 u^h(x,y)&=&b_1 +b_2x+b_3y \nn\\
@@ -47,13 +49,14 @@
 where $b_i$ are constants to be determined and which depend on the triangle shape.
 Note that the strain rate components are then given by
 \begin{eqnarray}
-\dot\varepsilon_{xx}&=&b_2  \nn\\
-\dot\varepsilon_{yy}&=&b_6  \nn\\
-\dot\varepsilon_{xy}&=&(b_3+b_5)/2 \nn
+\dot\varepsilon_{xx}(\vec\upnu)&=&b_2  \nn\\
+\dot\varepsilon_{yy}(\vec\upnu)&=&b_6  \nn\\
+\dot\varepsilon_{xy}(\vec\upnu)&=&(b_3+b_5)/2 \nn
 \end{eqnarray}
 and are constant throughout the element.
 
-The velocities should satisfy the following six equations:
+The velocities should satisfy the following six equations (when it is evaluated at a node we should 
+recover the nodal velocity):
 \begin{eqnarray}
 u_1 &=& u^h(x_1,y_1)= b_1 + b_2x_1+b_3y_1 \nn\\
 u_2 &=& u^h(x_2,y_2)= b_1 + b_2x_2+b_3y_2 \nn\\
@@ -87,7 +90,6 @@
 \]
 In order to obtain $b_1,b_2,b_3$ we need to solve this system, or simply to compute the
 inverse of the $3\times 3$ ${\bm M}$ matrix, as explained in Appendix~\ref{sec:inv3x3}.
-
 We define $D={\rm det}({\bm M})$ and we get
 \[
 \left(
@@ -122,7 +124,7 @@
 %\end{array}
 %\right)
 \]
-The matrix $\tilde{\bm M}$ writes:
+The matrix $\tilde{\bm M}$ is given by:
 \[
 \tilde{\bm M}
 %=
@@ -166,13 +168,12 @@
 \bN_2(x,y) &=& \frac{1}{D}[(x_3y_1-x_1y_3) + (y_3-y_1)x + (x_1-x_3)y] \nn\\
 \bN_3(x,y) &=& \frac{1}{D}[(x_1y_2-x_2y_1) + (y_1-y_2)x + (x_2-x_1)y] \nn
 \end{eqnarray}
-Importantly we can easily verify that for example
-\[
-\bN_2(x_1,y_1)= \frac{1}{D}[(x_3y_1-x_1y_3) + (y_3-y_1)x_1 + (x_1-x_3)y_1] = 0
-\]
-\[
-\bN_2(x_2,y_2)= \frac{1}{D}[(x_3y_1-x_1y_3) + (y_3-y_1)x_2 + (x_1-x_3)y_2] = 1
-\]
+We can then easily verify that for example
+\begin{eqnarray}
+\bN_2(x_1,y_1)&=& \frac{1}{D}[(x_3y_1-x_1y_3) + (y_3-y_1)x_1 + (x_1-x_3)y_1] = 0 \\
+\bN_2(x_2,y_2)&=& \frac{1}{D}[(x_3y_1-x_1y_3) + (y_3-y_1)x_2 + (x_1-x_3)y_2] = 1 \\
+\bN_2(x_3,y_3)&=& \frac{1}{D}[(x_3y_1-x_1y_3) + (y_3-y_1)x_3 + (x_1-x_3)y_3] = 0 
+\end{eqnarray}
 Note that the area $A$ of the triangle is given by:
 \[
 A=\frac{1}{2}D = \frac{1}{2}
@@ -201,7 +202,7 @@
 \bN_2(r,s) &=& s 
 \end{eqnarray}
 \end{mdframed}
-Once again we can verify that $\bN_i(x_j,y_j)=\delta_{ij}$ and $\sum_i \bN_i(r,s)=1$.
+Once again we can verify that $\bN_i(x_j,y_j)=\delta_{ij}$ and $\sum\limits_i \bN_i(r,s)=1$.
 
 
 

basis_Q1_2D.tex

@@ -13,7 +13,6 @@
 to the $(r,s)$ once and vice versa will be dealt with later on.
 The basis functions in the above reference element in the reduced 
 coordinates system $(r,s)$ are given by:
-
 \begin{mdframed}[backgroundcolor=blue!5]
 \begin{eqnarray}
 \bN_1(r,s)&=&0.25(1-r)(1-s) \nonumber\\
@@ -24,19 +23,15 @@
 \end{mdframed}
 These basis functions are the product of the linear basis functions of Section~\ref{sec:bf1}
 in the $r$ direction and the $s$ direction.
-
 \begin{center}
 \includegraphics[width=4cm]{images/basis_Q1_2D/N1}
 \includegraphics[width=4cm]{images/basis_Q1_2D/N2}
 \includegraphics[width=4cm]{images/basis_Q1_2D/N3}
 \includegraphics[width=4cm]{images/basis_Q1_2D/N4}\\
-{\captionfont Surface representation of the basis functions on the reference element.\\
+{\captionfont Surface representation of the basis functions on the reference element.
 {\color{gray} in images/basis\_Q1\_2D/ }}
 \end{center}
-
-
 The partial derivatives of these functions with respect to $r$ ans $s$ automatically follow:
-
 \begin{mdframed}[backgroundcolor=blue!5]
 \begin{align}
 \frac{\partial \bN_1}{\partial r}(r,s)&= - 0.25(1-s) &
@@ -49,7 +44,6 @@
 \frac{\partial \bN_4}{\partial s}(r,s)&= + 0.25(1-r) \nonumber
 \end{align}
 \end{mdframed}
-
 Let us go back to Eq.~\eqref{bf01} and let us assume that the 
 function $v(r,s)=C$ so that $v_i=C$ for $i=1,2,3,4$. 
 It then follows that 

basis_Q28_2D.tex

@@ -2,9 +2,8 @@
 %~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
 The serendipity elements are those rectangular elements which have no
-interior nodes \cite[p65]{reddybook2}.
-
-Inside an element the local numbering of the nodes is as follows:
+interior nodes (See for example Reddy \cite[p65]{reddybook2}).
+Inside an element a possible local numbering of the nodes is as follows:
 
 \input{tikz/tikz_serendipity2D}
 
@@ -44,7 +43,7 @@
 \includegraphics[width=4cm]{images/basis_Q28_2D/N6}
 \includegraphics[width=4cm]{images/basis_Q28_2D/N7}
 \includegraphics[width=4cm]{images/basis_Q28_2D/N8}\\
-{\captionfont Surface representation of the basis functions on the reference element.\\
+{\captionfont Surface representation of the basis functions on the reference element.
 {\color{gray} in images/basis\_Q28\_2D/ }}
 \end{center}
 
@@ -77,5 +76,4 @@
 \frac{\partial \bN_7}{\partial s}(r,s)&=&  (r-1)s
 \end{eqnarray}
 \end{mdframed}
-
 These basis functions are used in \stone 52.

basis_Q2_2D.tex

@@ -3,12 +3,12 @@
 
 This element is part of the so-called Lagrange family \cite{raki00}. 
 Inside an element the local numbering of the nodes is as follows\footnote{I have adopted here 
-a numbering scheme starting at zero!}:
+a numbering scheme starting at zero! Also, it is a numbering among many other possible choices!}:
 
 \input{tikz/tikz_q22d}
 
 Note that this numbering is also employed in Li \cite[p56]{li06}.
-The polynomial representation of the function $\phi$ over this element is then
+The polynomial representation of the function $\phi$ over this element is then taken to be biquadratic:
 \[
 \phi^h(r,s) = a + br + cs + drs + er^2 + fs^2 + gr^2s + hrs^2 + i r^2s^2 = \sum_{i=0}^8 \bN_i(r,s) \phi_i
 \]
@@ -26,8 +26,7 @@
 \bN_8(r,s)&=&     (1-r^2)      (1-s^2)\nonumber
 \end{eqnarray}
 \end{mdframed}
-
-Note that we have $\bN_i(r_j,s_j)=\delta_{ij}$ and $\bN_i(r_i,s_i)=1$. 
+Note that we have $\bN_i(r_j,s_j)=\delta_{ij}$ and then obviously $\bN_i(r_i,s_i)=1$. 
 
 \begin{center}
 \includegraphics[width=4cm]{images/basis_Q2_2D/N1}
@@ -39,11 +38,9 @@
 \includegraphics[width=4cm]{images/basis_Q2_2D/N7}
 \includegraphics[width=4cm]{images/basis_Q2_2D/N8}
 \includegraphics[width=4cm]{images/basis_Q2_2D/N9}\\
-{\captionfont Surface representation of the basis functions on the reference element.\\
+{\captionfont Surface representation of the basis functions on the reference element.
 {\color{gray} in images/basis\_Q2\_2D/ }}
 \end{center}
-
-
 Their derivatives are given by:
 \begin{mdframed}[backgroundcolor=blue!5]
 \begin{align}
@@ -67,5 +64,4 @@
 \frac{\partial \bN_8}{\partial s}&=     (1-r^2)        (-2s)\nonumber
 \end{align}
 \end{mdframed}
-
-These basis functions are used in \stone 18.
+These basis functions are used for example in \stone 18.

basis_Q3_2D.tex

@@ -1,38 +1,37 @@
 \begin{flushright} {\tiny {\color{gray} basis\_Q3\_2D.tex}} \end{flushright}
 %~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
-Inside an element the local numbering of the nodes is as follows:
+Inside an element a possible local numbering of the nodes is as follows:
 
 \input{tikz/tikz_q32d}
 
-The 1D cubic basis functions are given by:
+As shown in Section~\ref{sec:bf3} the 1D cubic basis functions are given by:
 \begin{align}
-\bN_1(r)&=(-1   +r +9r^2 - 9r^3)/16 & \bN_1(t)&=(-1   +t +9t^2 - 9t^3)/16 \nonumber\\
-\bN_2(r)&=(+9 -27r -9r^2 +27r^3)/16 & \bN_2(t)&=(+9 -27t -9t^2 +27t^3)/16 \nonumber\\
-\bN_3(r)&=(+9 +27r -9r^2 -27r^3)/16 & \bN_3(t)&=(+9 +27t -9t^2 -27t^3)/16 \nonumber\\
-\bN_4(r)&=(-1   -r +9r^2 + 9r^3)/16 & \bN_4(t)&=(-1   -t +9t^2 + 9t^3)/16 \nonumber
+\bN_1(r)&=(-1   +r +9r^2 - 9r^3)/16 & \bN_1(s)&=(-1   +s +9s^2 - 9s^3)/16 \nonumber\\
+\bN_2(r)&=(+9 -27r -9r^2 +27r^3)/16 & \bN_2(s)&=(+9 -27s -9s^2 +27s^3)/16 \nonumber\\
+\bN_3(r)&=(+9 +27r -9r^2 -27r^3)/16 & \bN_3(s)&=(+9 +27s -9s^2 -27s^3)/16 \nonumber\\
+\bN_4(r)&=(-1   -r +9r^2 + 9r^3)/16 & \bN_4(s)&=(-1   -s +9s^2 + 9s^3)/16 \nonumber
 \end{align}
-
-The resulting basis functions are simply the tensor product of the above 1D ones:
+and the resulting 2D basis functions are simply the tensor product of the above 1D ones:
 
 \begin{mdframed}[backgroundcolor=blue!5]
 \begin{eqnarray}
-\bN_{01}(r,s)&=&\bN_1(r)\bN_1(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (-1  +t +9s^2 - 9s^3)/16 \nonumber\\
-\bN_{02}(r,s)&=&\bN_2(r)\bN_1(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (-1  +t +9s^2 - 9s^3)/16 \nonumber\\
-\bN_{03}(r,s)&=&\bN_3(r)\bN_1(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (-1  +t +9s^2 - 9s^3)/16 \nonumber\\
-\bN_{04}(r,s)&=&\bN_4(r)\bN_1(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (-1  +t +9s^2 - 9s^3)/16 \nonumber\\
+\bN_{01}(r,s)&=&\bN_1(r)\bN_1(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (-1  +s +9s^2 - 9s^3)/16 \nonumber\\
+\bN_{02}(r,s)&=&\bN_2(r)\bN_1(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (-1  +s +9s^2 - 9s^3)/16 \nonumber\\
+\bN_{03}(r,s)&=&\bN_3(r)\bN_1(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (-1  +s +9s^2 - 9s^3)/16 \nonumber\\
+\bN_{04}(r,s)&=&\bN_4(r)\bN_1(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (-1  +s +9s^2 - 9s^3)/16 \nonumber\\
 \bN_{05}(r,s)&=&\bN_1(r)\bN_2(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (9 -27s -9s^2 +27s^3)/16 \nonumber\\
 \bN_{06}(r,s)&=&\bN_2(r)\bN_2(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (9 -27s -9s^2 +27s^3)/16 \nonumber\\
 \bN_{07}(r,s)&=&\bN_3(r)\bN_2(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (9 -27s -9s^2 +27s^3)/16 \nonumber\\
 \bN_{08}(r,s)&=&\bN_4(r)\bN_2(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (9 -27s -9s^2 +27s^3)/16 \nonumber\\
-\bN_{09}(r,s)&=&\bN_1(r)\bN_3(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (9 +27t -9t^2 -27t^3)/16 \nn\\
-\bN_{10}(r,s)&=&\bN_2(r)\bN_3(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (9 +27t -9t^2 -27t^3)/16 \nn\\
-\bN_{11}(r,s)&=&\bN_3(r)\bN_3(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (9 +27t -9t^2 -27t^3)/16 \nn\\
-\bN_{12}(r,s)&=&\bN_4(r)\bN_3(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (9 +27t -9t^2 -27t^3)/16 \nn\\
-\bN_{13}(r,s)&=&\bN_1(r)\bN_4(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (-1   -t +9t^2 + 9t^3)/16\nn\\
-\bN_{14}(r,s)&=&\bN_2(r)\bN_4(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (-1   -t +9t^2 + 9t^3)/16\nn\\
-\bN_{15}(r,s)&=&\bN_3(r)\bN_4(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (-1   -t +9t^2 + 9t^3)/16\nn\\
-\bN_{16}(r,s)&=&\bN_4(r)\bN_4(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (-1   -t +9t^2 + 9t^3)/16
+\bN_{09}(r,s)&=&\bN_1(r)\bN_3(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (9 +27s -9s^2 -27s^3)/16 \nn\\
+\bN_{10}(r,s)&=&\bN_2(r)\bN_3(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (9 +27s -9s^2 -27s^3)/16 \nn\\
+\bN_{11}(r,s)&=&\bN_3(r)\bN_3(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (9 +27s -9s^2 -27s^3)/16 \nn\\
+\bN_{12}(r,s)&=&\bN_4(r)\bN_3(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (9 +27s -9s^2 -27s^3)/16 \nn\\
+\bN_{13}(r,s)&=&\bN_1(r)\bN_4(s) = (-1   +r +9r^2 - 9r^3)/16 \cdot (-1   -s +9s^2 + 9s^3)/16\nn\\
+\bN_{14}(r,s)&=&\bN_2(r)\bN_4(s) = (+9 -27r -9r^2 +27r^3)/16 \cdot (-1   -s +9s^2 + 9s^3)/16\nn\\
+\bN_{15}(r,s)&=&\bN_3(r)\bN_4(s) = (+9 +27r -9r^2 -27r^3)/16 \cdot (-1   -s +9s^2 + 9s^3)/16\nn\\
+\bN_{16}(r,s)&=&\bN_4(r)\bN_4(s) = (-1   -r +9r^2 + 9r^3)/16 \cdot (-1   -s +9s^2 + 9s^3)/16
 \end{eqnarray}
 \end{mdframed}
 
@@ -53,18 +52,13 @@
 \includegraphics[width=4cm]{images/basis_Q3_2D/N14}
 \includegraphics[width=4cm]{images/basis_Q3_2D/N15}
 \includegraphics[width=4cm]{images/basis_Q3_2D/N16}\\
-{\captionfont Surface representation of the basis functions on the reference element.\\
+{\captionfont Surface representation of the basis functions on the reference element.
 {\color{gray} in images/basis\_Q3\_2D/ }}
 \end{center}
-
 The derivatives are trivial to obtain from the derivatives of the 1D basis functions, 
 e.g.
 \[
 \frac{\partial \bN_{13}}{\partial r} = 
-\frac{\partial \bN_{1}}{\partial r} \bN_4(s) 
+\frac{\partial \bN_{1}}{\partial r} \bN_3(s) 
 \]
-
 These basis functions are used in \stone 19.
-
-
-

basis_QH8_2D.tex

@@ -0,0 +1,74 @@
+
+This element is proposed in Zhang \& Xiang (2020) \cite{zhxi20}. Two remarks
+must be made: 1) Eq.~(29) of their publication which is the definition
+of the basis functions contains an error\footnote{
+Answer from the author: ``N5 to N8 is missing an A in the denominator and 
+the calculation program does not have this problem''}. 
+2) The authors use a rather 
+uncommon and annoying rotated numbering:
+\begin{verbatim}
+      y
+      |
+2=====5=====1             3=====6=====2
+|           |             |           |   (r_0,s_0)=(-1,-1)   (r_4,s_4)=( 0,-1)
+|           |             |           |   (r_1,s_1)=(+1,-1)   (r_5,s_5)=(+1, 0)
+6           8--x          7     +     5   (r_2,s_2)=(+1,+1)   (r_6,s_6)=( 0,+1)
+|           |             |           |   (r_3,s_3)=(-1,+1)   (r_7,s_7)=(-1, 0)
+|           |             |           |    
+3=====7=====4             0=====4=====1
+Zhang & Xiang             our numbering
+\end{verbatim}
+
+For each element they define (their numbering):
+\begin{eqnarray}
+A   &=& \frac{1}{2} [ (x_1-x_3)(y_2-y_4)-(x_2-x_4)(y_1-y_3) ] \nn\\
+m_x &=& (x_1-x_4)(y_2-y_3)-(x_2-x_3)(y_1-y_4) \nn\\
+m_y &=& (x_3-x_4)(y_1-y_2)-(x_1-x_2)(y_3-y_4) \nn
+\end{eqnarray}
+Note that $A$ is the area of the element, and that in the case when 
+the element is a rectangle then $m_x=m_y=0$.
+\begin{eqnarray}
+\bN_1(r,s)&=& n_1(r,s) +(m_x^2 - m_xm_y + m_y^2)\frac{E(r,s)}{D} \nn\\
+\bN_2(r,s)&=& n_2(r,s) +(m_x^2 + m_xm_y + m_y^2)\frac{E(r,s)}{D} \nn\\
+\bN_3(r,s)&=& n_3(r,s) +(m_x^2 - m_xm_y + m_y^2)\frac{E(r,s)}{D} \nn\\
+\bN_4(r,s)&=& n_4(r,s) +(m_x^2 + m_xm_y + m_y^2)\frac{E(r,s)}{D} \nn\\
+\bN_5(r,s)&=& n_5(r,s) -m_x(2Am_x+m_y^2)\frac{E(r,s)}{AD} \nn\\
+\bN_6(r,s)&=& n_6(r,s) -m_y(2Am_y+m_x^2)\frac{E(r,s)}{AD} \nn\\
+\bN_7(r,s)&=& n_7(r,s) +m_x(-2Am_x+m_y^2)\frac{E(r,s)}{AD} \nn\\
+\bN_8(r,s)&=& n_8(r,s) +m_y(-2Am_y+m_x^2)\frac{E(r,s)}{AD} \nn
+\end{eqnarray}
+with 
+\[
+E(r,s)=(1-r^2)(1-s^2)
+\qquad
+D=4(4A^2+m_x^2+m_y^2)
+\]
+and where the $n_i$ functions are the basis functions of the 'regular' 
+8-node element (see Section~\ref{sec:serendipity2D}).
+
+This is implemented in \stone 52.
+
+\todo[inline]{not finished. SHOW CONSISTENCY !! like in paper
+email sent to author about mistake.  }
+
+Let us verify consistency:
+\begin{eqnarray}
+\sum_{i=1}^8 \bN_i(r,s) 
+&=& \underbrace{\sum_{i=1}^8 n_i(r,s)}_{=0} + \frac{E(r,s)}{D} 
+\left[
+(m_x^2 - m_xm_y + m_y^2)
++(m_x^2 + m_xm_y + m_y^2)
++(m_x^2 - m_xm_y + m_y^2)
++(m_x^2 + m_xm_y + m_y^2) \right. \nn\\
+&& \left.
+-m_x(2Am_x+m_y^2)\frac{1}{A}
+-m_y(2Am_y+m_x^2)\frac{1}{A}
++m_x(-2Am_x+m_y^2)\frac{1}{A}
++m_y(-2Am_y+m_x^2)\frac{1}{A}
+\right] \nn\\
+&=& \frac{E(r,s)}{D}
+\left[(4m_x^2 + 4m_y^2) + \frac{1}{A} (-4Am_x^2 - 4A m_y^2) \right]  \\
+&=& 0
+\end{eqnarray}
+
+

basis_functions_meaning.tex

@@ -12,7 +12,7 @@ \subsubsection{In one dimension}
 
 The four cases a,b,c,d are examples of combinations of these basis functions:
 \[
-f_h(x)=\sum_{i=1}^4 \bN_i(x) f_i
+f^h(x)=\sum_{i=1}^4 \bN_i(x) f_i
 \]
 Where $f_i$ are the values associated to the four nodes. 
 We assume that the distance $h$ between nodes is 1.
@@ -25,15 +25,15 @@ \subsubsection{In one dimension}
 
 Example c) illustrates the fact that these linear basis functions can exactly represent a
 linear function. When $f(x)=x-2$, then $f_1=f(0)=-2$, $f_2=f(1)=-1$, $f_3=f(2)=0$ and $f_4=f(3)=+1$, 
-then $f_h(x)$ is exactly $f(x)$ on the domain.
+then $f^h(x)$ is exactly $f(x)$ on the domain.
 
 Example d) illustrates the fact that linear basis functions cannot represent a parabola. Smaller and 
 smaller elements will do an increasingly better job and will get closer to the curve but a 
 systematic error will subsist.  
 
 
 Note that these drawings are trivial to produce since $\bN_i(x_j)=\delta_{ij}$ by definition, so that 
-$f_h(x_j)=f_j$.
+$f^h(x_j)=f_j$.
 
 %........................................................................
 \subsubsection{In two dimensions}